I don't know of a reference, but here is a quick argument. Suppose we want to compute the homotopy pullback P = X ×hZ Y of two maps f : X → Z and g : Y → Z of pointed simplicial sets. Assume for convenience that everything is fibrant. There is a fibration ZΔ[1] → Z∂Δ[1] = Z × Z with fiber ΩZ. Now P is the pullback of the diagram X × Y → Z × Z ← ZΔ[1]. In particular, P → X × Y is also a fibration with fiber ΩZ, and the Mayer-Vietoris sequence follows from the long exact sequence of homotopy groups of this fibration.
I don't think the Seifert-van Kampen theorem follows from these kinds of considerations. Rather, it is the statement that the fundamental groupoid functor $\tau_{\leq 1}$ preserves homotopy colimits. That's because it is left adjoint to the inclusion of 1-groupoids in $\infty$-groupoids (a generalization to any $\infty$-category is in Higher Topos Theory, Prop. 5.5.6.18).
[Removed the part on cohomology because it got me confused!]
Added:
I attempted to explain the long exact sequences in cohomology without using spectra, but that was wrong. Here's a correct explanation. The reduced cohomology of a pointed space $X$ depends only on its stabilization $\Sigma^\infty X$: it is given by $H^n(X;A)=[\Sigma^\infty X,\Sigma^n HA]$ where $HA$ is an infinite delooping of $K(A,0)$. The functor $\Sigma^\infty$ preserves cofiber sequences (being left adjoint). Now if you have a cofiber sequence in a stable category, you get long exact sequences of abelian groups when you apply functors like $[E,-]$ or $[-,E]$.
Added later:
My original answer was correct, but like I said I got confused… Here it is again. Let $A\to B\to C$ be a cofiber sequence of pointed spaces. As you say in your question, you get a fiber sequence of mapping spaces
$Map(C,X)\to Map(B,X)\to Map(A,X)$
for any $X$, because $Map$ transforms homotopy colimits in its first variable into homotopy limits. In its second variable it preserves homotopy limits, so $\Omega Map(A,X)=Map(A,\Omega X)$. Applying to $X=K(G,n)$ gives you the usual long exact sequence in cohomology, but only from $H^0$ to $H^n$.
Best Answer
Mayer-Vietoris sequences can be obtained from excision isomorphisms.
Anything worthy of the name "homology theory" will give a long exact sequence $$\dots \to h_n(A)\to h_n(X)\to h_n(A\to X)\to h_{n-1}(A)\to \dots$$ for each morphism $A\to X$. And for a square, a.k.a. map of morphisms $(C\to B)\to (A\to X)$, it will give you a map from the long exact sequence of $C\to B$ to that of $A\to X$.
If the square happens to induce an isomorphism $h_n(C\to B)\to h_n(A\to X)$ for every $n$ (as it will if it is a homotopy pushout square) then the desired map $h_n(X)\to h_{n-1}(C)$ is given by composing $h_n(X)\to h_n(A\to X)\cong h_n(C\to B)\to h_{n-1}(C)$, and a little diagram chase gives you the exactness you want.
Depending on how one axiomatizes the notion of homology theory, the fact that homotopy pushout squares gives isomorphisms in relative homology is either an axiom or a consequence of the axioms.
There is no need to work with spectra or other stable objects here.
Note that if you had something like a homology theory except that it gave isomorphisms $h_n(C\to B)\to h_n(A\to X)$ for homotopy pullbacks instead of for homotopy pushouts then you would get a "Mayer-Vietoris sequence" for pullback squares. That's how it is for homotopy groups and based spaces, except that things get funny down around $\pi_1$ and relative $\pi_2$.